b) Ta có
$x^4 + 2019x^2 + 2018x + 2019= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2019x^2 + 2019x + 2019$
$= x^2(x^2+x+1) - x(x^2+x+1) + 2019(x^2+x+1)$
$= (x^2+x+1)(x^2-x+2019)$
Vậy $x^4 + 2019x^2 + 2018x + 2019 = (x^2+x+1)(x^2-x+2019)$.
d) Ta có
$x^4 + 64 = x^4 + 16x^2 + 64 -16x^2$
$= (x^2)^2 + 2.x.8 + 8^2 -16x^2$
$= (x^2+8)^2 - (4x)^2$
$= (x^2-4x+8)(x^2+4x+8)$
Vậy $x^4 + 64 = (x^2-4x+8)(x^2+4x+8)$.
